Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, beat[101][101];
int try1[101][101], coloane[101], linii[101], t1;
int try2[101][101], colums[101], rows[101], t2;
bool possible1, possible2;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> beat[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    linii[i] = 501;
    for (int j = 1; j <= m; j++) {
      linii[i] = min(linii[i], beat[i][j]);
    }
    t1 += linii[i];
    for (int j = 1; j <= m; j++) {
      try1[i][j] += linii[i];
    }
  }
  for (int j = 1; j <= m; j++) {
    coloane[j] = beat[1][j] - try1[1][j];
    ;
    t1 += coloane[j];
  }
  for (int j = 1; j <= m; j++) {
    for (int i = 1; i <= n; i++) {
      try1[i][j] += coloane[j];
    }
  }
  for (int j = 1; j <= m; j++) {
    colums[j] = 501;
    for (int i = 1; i <= n; i++) {
      colums[j] = min(colums[j], beat[i][j]);
    }
    t2 += colums[j];
    for (int i = 1; i <= n; i++) {
      try2[i][j] += colums[j];
    }
  }
  for (int i = 1; i <= n; i++) {
    rows[i] = beat[i][1] - try2[i][1];
    t2 += rows[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      try2[i][j] += rows[i];
    }
  }
  possible1 = true;
  possible2 = true;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (beat[i][j] != try1[i][j]) {
        possible1 = false;
      }
      if (beat[i][j] != try2[i][j]) {
        possible2 = false;
      }
    }
  }
  if (!possible1 && !possible2) {
    cout << "-1";
  } else if (!possible1 && possible2) {
    cout << t2 << '\n';
    for (int i = 1; i <= n; i++) {
      while (rows[i]) {
        cout << "row"
             << " " << i << '\n';
        rows[i]--;
      }
    }
    for (int j = 1; j <= m; j++) {
      while (colums[j]) {
        cout << "col"
             << " " << j << '\n';
        colums[j]--;
      }
    }
  } else if (possible1 && !possible2) {
    cout << t1 << '\n';
    for (int i = 1; i <= n; i++) {
      while (linii[i]) {
        cout << "row"
             << " " << i << '\n';
        linii[i]--;
      }
    }
    for (int j = 1; j <= m; j++) {
      while (coloane[j]) {
        cout << "col"
             << " " << j << '\n';
        coloane[j]--;
      }
    }
  } else if (possible1 && possible2) {
    if (t1 > t2) {
      cout << t2 << '\n';
      for (int i = 1; i <= n; i++) {
        while (rows[i]) {
          cout << "row"
               << " " << i << '\n';
          rows[i]--;
        }
      }
      for (int j = 1; j <= m; j++) {
        while (colums[j]) {
          cout << "col"
               << " " << j << '\n';
          colums[j]--;
        }
      }
    } else {
      cout << t1 << '\n';
      for (int i = 1; i <= n; i++) {
        while (linii[i]) {
          cout << "row"
               << " " << i << '\n';
          linii[i]--;
        }
      }
      for (int j = 1; j <= m; j++) {
        while (coloane[j]) {
          cout << "col"
               << " " << j << '\n';
          coloane[j]--;
        }
      }
    }
  }
}